Moderate -0.5 This is a straightforward modulus inequality requiring students to consider two cases (x ≥ 2 and x < 2) and solve linear inequalities in each case. While it requires understanding of the modulus function definition, the algebraic manipulation is simple and this is a standard textbook exercise with no novel insight required, making it slightly easier than average.
EITHER: State or imply non-modular inequality \((x-2)^2 > (2x-3)^2\), or corresponding equation
B1
Solve a 3-term quadratic, as in Q1
M1
Obtain critical value \(x = \frac{5}{3}\)
A1
State final answer \(x < \frac{5}{3}\) only
A1
OR1: State the relevant critical linear inequality \((2-x) > (2x-3)\), or corresponding equation
B1
Solve inequality or equation for \(x\)
M1
Obtain critical value \(x = \frac{5}{3}\)
A1
State final answer \(x < \frac{5}{3}\) only
A1
OR2: Make recognisable sketches of \(y = 2x-3\) and \(y =
x-2
\) on a single diagram
Find x-coordinate of the intersection
M1
Obtain \(x = \frac{5}{3}\)
A1
State final answer \(x < \frac{5}{3}\) only
A1
4
**EITHER:** State or imply non-modular inequality $(x-2)^2 > (2x-3)^2$, or corresponding equation | B1 |
Solve a 3-term quadratic, as in Q1 | M1 |
Obtain critical value $x = \frac{5}{3}$ | A1 |
State final answer $x < \frac{5}{3}$ only | A1 |
**OR1:** State the relevant critical linear inequality $(2-x) > (2x-3)$, or corresponding equation | B1 |
Solve inequality or equation for $x$ | M1 |
Obtain critical value $x = \frac{5}{3}$ | A1 |
State final answer $x < \frac{5}{3}$ only | A1 |
**OR2:** Make recognisable sketches of $y = 2x-3$ and $y = |x-2|$ on a single diagram | B1 |
Find x-coordinate of the intersection | M1 |
Obtain $x = \frac{5}{3}$ | A1 |
State final answer $x < \frac{5}{3}$ only | A1 | 4 |